Buck Boost Transformer Theory

Status
Not open for further replies.
D

Dustin Foelber

Senior Member
I am a bit confused by the sizing of buck boost transformer. I see that a 1KVA transformer can be used to boost 208V single phase to 240V and provide nearly 30A of operating load. Can someone explain the theory behind why the transformer does not need to be rated for the amperage of the load even though it passes through this transformer and both the primary and secondary of its windings
 
The 30amp load only passes through the secondary. The primary connection drives the transformer to make the boost/buck voltage and only needs enough power to make 30 amps at that voltage (e.g. 32 volts out @ 30 amp = 960va = 4.6 amps in at 208 volt, not counting losses).

You could drive the B/B transformer primary with a completely different circuit from the actual load but I don't recall if the code allows that; I wouldn't on general principles. (Too lazy to look this up)
 
Without going into the detail and the equations, you may find some help in this single point:

We'll look at the simple case of a single phase 208 to 240 transformer, made up of a single 208V to 32V transformer.
Primary winding AB and secondary winding CD. You make an autotransformer by connecting B to C. You can then think of this as one winding, A (BC) D.
You connect the incoming 208 to A and (BC) and take the output between A and D. If the full 30A load current flowed through A (BC) D, the transformer would have to be rated for at least 240V x 30A.
But if you look at the actual current flow, the 30A going to the load comes in at (BC) and passes though the larger guage winding only to get to D. The current through A (BC) is only the lower current needed to allow 30A to pass through (BC) D.
This means that the power rating of the transformer only needs to be 32V x 30A.
 
GoldDigger said:
Without going into the detail and the equations, you may find some help in this single point:

We'll look at the simple case of a single phase 208 to 240 transformer, made up of a single 208V to 32V transformer.
Primary winding AB and secondary winding CD. You make an autotransformer by connecting B to C. You can then think of this as one winding, A (BC) D.
You connect the incoming 208 to A and (BC) and take the output between A and D. If the full 30A load current flowed through A (BC) D, the transformer would have to be rated for at least 240V x 30A.
But if you look at the actual current flow, the 30A going to the load comes in at (BC) and passes though the larger guage winding only to get to D. The current through A (BC) is only the lower current needed to allow 30A to pass through (BC) D.
This means that the power rating of the transformer only needs to be 32V x 30A.
Click to expand...
Got it. Appreciate the feedback. What if I have neutral? Do I pass that through the transformer? How does it know to find the center tap to properly give you 120 between phases?
 
Dustin Foelber said:
Got it. Appreciate the feedback. What if I have neutral? Do I pass that through the transformer? How does it know to find the center tap to properly give you 120 between phases?
Click to expand...
My example was based on a single phase 208V line to neutral source. And a single boost transformer.
If you bring in all three (or two out of three) phases boost function gets more complicated.
If you start with a 208Y/120 source, the easiest boost configuration to understand is to use three 120 to 139 volt boost transformers to get a 240Y/139 source with the same central point as the neutral.
If you use a boost transformer connected line to line to get the 208V input, you will get a 240V output where one end still has 120V to the neutral and the other does not. (I have not done the vector analysis to see just what that voltage is.
If you start with a 208V delta, with no neutral or ignoring the neutral, you could use two boost transformers to get an assymmetric open delta 240 or you can use three boost transformers to get a symmetric closed delta 240V in which no output leads is common with an input lead.
You cannot use a boost transformer on a 208V wye to get a 120-0-120 source with a neutral. You will have to use an isolation transformer so that you can ground the center tap of the output. And that transfomer will need to be fully rated for the load.
 
For the voltage reduction, the video showed him using the secondary to actually do subtractive bucking, but I've always seen it done by placing the secondary in series with the input instead of the output, like this:

1670293257170.png
 
I think the main practical difference between the two is that for subtractive buck the primary needs to be rated for 240, while in your example it only needs to be rated for 208.
 
zbang said:
You could drive the B/B transformer primary with a completely different circuit from the actual load but I don't recall if the code allows that; I wouldn't on general principles. (Too lazy to look this up)
Click to expand...
I believe the two circuits would need to be in phase with one another to, if anything would create excessive current and heating if not in phase.

Code wise would probably at least require more poles and/or handle ties on switching/disconnecting means
 
kwired said:
I believe the two circuits would need to be in phase with one another to, if anything would create excessive current and heating if not in phase.

Code wise would probably at least require more poles and/or handle ties on switching/disconnecting means
Click to expand...
There will be no excessive current or heating if the two sources are not precisely in phase. The bucked or boosted voltage will simply be to vector sum of the transformer secondary voltage and the original voltage rather than the scalar sum or difference. So the result may not be what was intended.
And if a second source circuit is used, the resulting equivalent circuit will not be an autotransformer.
 
zbang said:
You could drive the B/B transformer primary with a completely different circuit from the actual load but I don't recall if the code allows that; I wouldn't on general principles. (Too lazy to look this up)
Click to expand...
kwired said:
... Code wise would probably at least require more poles and/or handle ties on switching/disconnecting means
Click to expand...

450.4(A) says "An over-current device shall not be installed in series with the shunt winding (the winding common to both the input and output circuits) of the autotransformer ..."
I suspect that's because if the shunt winding is open-circuited, then there could be relatively high voltage spikes on this winding when the core is in the linear region between alternating polarities of magnetic saturation. This will occur because the secondary of a buck-boost will be driven due to its location in series between the line input and load. With a buck-boost the "shunt winding" mentioned is the primary, and so a relatively high voltage can be developed across it due to the large transformation ratio going backwards from secondary to primary.

If you supply the primary of a buck-boost with a different circuit having its own over-current device, then I think high voltage spikes could also occur if that over-current device opens up. Having the two circuits be on different poles of the same breaker would help, but there still might be a high voltage spike if the pole contacts for the buck-boost primary open slightly before the other pole does.
 
GoldDigger said:
There will be no excessive current or heating if the two sources are not precisely in phase. The bucked or boosted voltage will simply be to vector sum of the transformer secondary voltage and the original voltage rather than the scalar sum or difference. So the result may not be what was intended.
And if a second source circuit is used, the resulting equivalent circuit will not be an autotransformer.
Click to expand...
That is true.
 
Status
Not open for further replies.
Top